GATE 2023 EE Q.56: solution and answer

GATE 2023 · EEQ.56 · 2 marks · NAT

Neglecting the delays due to the logic gates in the circuit shown in figure, the decimal equivalent of the binary sequence [ABCD] of initial logic states, which will not change with clock, is ________.

GATE 2023 EE question 56: Two D flip-flops, FF1 (left) and FF2 (right), both driven by a common CLK line. — Figure from the GATE 2023 EE paper. It is part of the question.

Answer and solution

ANSWER

The only state the clock leaves unchanged is $Q_{1} Q_{2} = 00$ . In that state $[A B C D] = (1000)_{2}$ , which is $8$ in decimal.

Official answer: 8Source: Official answer key, GATE 2023 EE (IIT Kanpur)

Read the nodes off the figure (complement written with ). Call the flip-flop outputs $Q_{1}$ (left flip-flop) and $Q_{2}$ (right flip-flop). Node A is wired to the left flip-flop's $\overline{Q}$ output, so $A = \overline{Q_{1}}$ . Both inputs of the NOR gate are tied to A, so it works as an inverter: $B = \overline{A + A} = \overline{A} = Q_{1}$ . Node C is on the wire from the right flip-flop's $Q$ output, so $C = Q_{2}$ . The X-OR output is node D, so $D = B \oplus C = Q_{1} \oplus Q_{2}$ .
Next-state equations: a D flip-flop copies its D input on the clock edge ( $Q^{+} = D$ ). The left flip-flop's D input is wired to $Q_{2}$ , and the right flip-flop's D input is wired to node D. So $Q_{1}^{+} = Q_{2}, Q_{2}^{+} = Q_{1} \oplus Q_{2}$
A state that does not change with the clock must satisfy $Q_{1}^{+} = Q_{1}$ and $Q_{2}^{+} = Q_{2}$ . That gives $Q_{1} = Q_{2}$ and $Q_{2} = Q_{1} \oplus Q_{2}$ .
Substitute $Q_{1} = Q_{2}$ into the second condition: $Q_{1} \oplus Q_{2} = 0$ , so $Q_{2} = 0$ and therefore $Q_{1} = 0$ . The only fixed state is $Q_{1} Q_{2} = 00$ .
Check with the state table ( $Q_{1} Q_{2} \to Q_{1}^{+} Q_{2}^{+}$ ): $00 \to 00$ , while $01 \to 11 \to 10 \to 01$ is a 3-state cycle. So $00$ is the only state that stays the same.
Node values in that state: $A = \overline{Q_{1}} = \overline{0} = 1$ , $B = Q_{1} = 0$ , $C = Q_{2} = 0$ , $D = 0 \oplus 0 = 0$ .
$[A B C D] = (1000)_{2} = 8$ .

KEY IDEA

A state that does not change with the clock is a fixed point of the next-state equations: for every D flip-flop, $Q^{+} = D$ must equal the present $Q$ . Write the D inputs in terms of $Q_{1}, Q_{2}$ , solve $Q_{1}^{+} = Q_{1}$ and $Q_{2}^{+} = Q_{2}$ , then work out the node values A, B, C, D from that state.

WATCH OUT FOR

Reading A as $Q_{1}$ instead of $\overline{Q_{1}}$ (A is wired to the $\overline{Q}$ output). That changes the equations to $Q_{2}^{+} = \overline{Q_{1}} \oplus Q_{2}$ , which gives the fixed state $11$ and the wrong answer $(1011)_{2} = 11$ . Two other traps: answering with the flip-flop state $Q_{1} Q_{2} = 00$ (giving 0) when the question asks for the four node values [ABCD], and forgetting that a NOR with both inputs tied together is just an inverter, so $B = \overline{A}$ .

From the GATE 2023 EE paper. Every question of this paper · Download the full paper · All 30 Sequential questions, with notes

GATE 2023 EE Q.56: solution and answer

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